Step 1

Product Rule of differentiation:

If f(x) and g(x) are any two differentiable functions, then

\(\displaystyle{f{{\left({g}\right)}}}'{\left({x}\right)}={f{{\left({x}\right)}}}\times{g}'{\left({x}\right)}+{f}'{\left({x}\right)}\times{g{{\left({x}\right)}}}\)

Chain rule of differentiation:

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{f{{\left({g{{\left({x}\right)}}}\right)}}}={f}'{\left({g{{\left({x}\right)}}}\right)}\times{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{g{{\left({x}\right)}}}\)

\(\displaystyle={f}'{\left({g{{\left({x}\right)}}}\right)}\times{g}'{\left({x}\right)}\times{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({x}\right)}\)

\(\displaystyle={f}'{\left({g{{\left({x}\right)}}}\right)}\times{g}'{\left({x}\right)}\)

The given function is \(\displaystyle{h}{\left({x}\right)}={\sin{{2}}}{x}{\cos{{2}}}{x}\)

Step 2

Differentiate the given function using the product rule as follows.

\(\displaystyle{h}'{\left({x}\right)}={\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\sin{{2}}}{x}{\cos{{2}}}{x}\right)}\)

\(\displaystyle={\sin{{2}}}{x}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\cos{{2}}}{x}\right)}+{\cos{{2}}}{x}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\sin{{2}}}{x}\right)}\)

(U sin g product rule)

\(\displaystyle={\sin{{2}}}{x}{\left(-{\sin{{2}}}{x}\right)}{\left({2}\right)}+{\cos{{2}}}{x}{\left({\cos{{2}}}{x}\right)}{\left({2}\right)}\)

(U sin g chain rule)

\(\displaystyle=-{2}{{\sin}^{{{2}}}{2}}{x}+{2}{{\cos}^{{{2}}}{2}}{x}\)

\(\displaystyle=-{2}{\left({1}-{{\cos}^{{{2}}}{2}}{x}\right)}+{2}{{\cos}^{{{2}}}{2}}{x}\)

\(\displaystyle=-{2}+{4}{{\cos}^{{{2}}}{2}}{x}\)

Therefore, the derivative of the given function is

\(\displaystyle{h}'{\left({x}\right)}=-{2}+{4}{{\cos}^{{{2}}}{2}}{x}\)

Product Rule of differentiation:

If f(x) and g(x) are any two differentiable functions, then

\(\displaystyle{f{{\left({g}\right)}}}'{\left({x}\right)}={f{{\left({x}\right)}}}\times{g}'{\left({x}\right)}+{f}'{\left({x}\right)}\times{g{{\left({x}\right)}}}\)

Chain rule of differentiation:

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{f{{\left({g{{\left({x}\right)}}}\right)}}}={f}'{\left({g{{\left({x}\right)}}}\right)}\times{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{g{{\left({x}\right)}}}\)

\(\displaystyle={f}'{\left({g{{\left({x}\right)}}}\right)}\times{g}'{\left({x}\right)}\times{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({x}\right)}\)

\(\displaystyle={f}'{\left({g{{\left({x}\right)}}}\right)}\times{g}'{\left({x}\right)}\)

The given function is \(\displaystyle{h}{\left({x}\right)}={\sin{{2}}}{x}{\cos{{2}}}{x}\)

Step 2

Differentiate the given function using the product rule as follows.

\(\displaystyle{h}'{\left({x}\right)}={\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\sin{{2}}}{x}{\cos{{2}}}{x}\right)}\)

\(\displaystyle={\sin{{2}}}{x}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\cos{{2}}}{x}\right)}+{\cos{{2}}}{x}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\sin{{2}}}{x}\right)}\)

(U sin g product rule)

\(\displaystyle={\sin{{2}}}{x}{\left(-{\sin{{2}}}{x}\right)}{\left({2}\right)}+{\cos{{2}}}{x}{\left({\cos{{2}}}{x}\right)}{\left({2}\right)}\)

(U sin g chain rule)

\(\displaystyle=-{2}{{\sin}^{{{2}}}{2}}{x}+{2}{{\cos}^{{{2}}}{2}}{x}\)

\(\displaystyle=-{2}{\left({1}-{{\cos}^{{{2}}}{2}}{x}\right)}+{2}{{\cos}^{{{2}}}{2}}{x}\)

\(\displaystyle=-{2}+{4}{{\cos}^{{{2}}}{2}}{x}\)

Therefore, the derivative of the given function is

\(\displaystyle{h}'{\left({x}\right)}=-{2}+{4}{{\cos}^{{{2}}}{2}}{x}\)